2021.Q2: Time Reversal and the Maximum of Brownian Motion
2021 Probability Prelim Exam (PDF)
Introduction
This problem tests several core ideas about Brownian motion that often appear implicitly in prelim problems:
- Understanding time-reversal of Brownian motion and recognizing when it produces another Brownian motion in distribution.
- Working comfortably with Gaussian processes, especially using mean and covariance to identify distributions.
- Connecting pathwise identities (almost sure statements) with distributional identities, especially involving extrema.
- Correctly invoking the reflection principle and understanding what it actually gives.
Main tools likely needed:
- Covariance structure of Brownian motion.
- Characterization of Gaussian processes by mean and covariance.
- Continuity of Brownian sample paths.
- Reflection principle for Brownian motion.
Pattern recognition cue:
If you see expressions like $B_{T-t} - B_T$, think “time reversal + centering,” and if you see maxima like $B_t^*$, expect the reflection principle to appear.
Problem Statement (verbatim)
Problem 2.
In this problem ${B_t, t \ge 0}$ represents a Standard Brownian Motion (SBM) defined on its canonical space
$(\Omega = C[0, \infty), {\mathcal{F}t}{t \ge 0}, \mathcal{F}, P)$.Let
\(B_t^* = \max_{0 \le s \le t} \{B_s\}, \quad t \ge 0\) (“trailing maximum” of SBM). Observe that with this notation the reflection principle states: \(B_t^* = |B_t| \quad \text{in distribution, for } t \ge 0.\) Let $T > 0$ be fixed.a. Let $Y_t = B_{T-t} - B_T,\; 0 \le t \le T.$
Calculate $\operatorname{Cov}(Y_t, Y_s),\; 0 \le t, s \le T.$b. Prove that ${Y_t, 0 \le t \le T} = {B_t, 0 \le t \le T}$ in distribution.
c.
(i) Prove that $Y_T^* = B_T^* - B_T,$ a.s.
(ii) Prove that $|B_T| = B_T^* - B_T$ in distribution.
Solutions
Part (a) Covariance computation
Claim.
For $0 \le s,t \le T$,
\(\operatorname{Cov}(Y_t, Y_s) = \min(s,t).\)
Proof.
By definition,
\(Y_t = B_{T-t} - B_T, \qquad Y_s = B_{T-s} - B_T.\)
Thus
\(\operatorname{Cov}(Y_t,Y_s)
= \operatorname{Cov}(B_{T-t}-B_T,\; B_{T-s}-B_T).\)
Expanding,
$$
\begin{aligned}
\operatorname{Cov}(Y_t,Y_s)
&= \operatorname{Cov}(B_{T-t},B_{T-s})
- \operatorname{Cov}(B_{T-t},B_T)
&\quad - \operatorname{Cov}(B_{T-s},B_T) - \operatorname{Cov}(B_T,B_T). \end{aligned} \(Using $\operatorname{Cov}(B_a,B_b)=a\wedge b$,\) \begin{aligned} \operatorname{Cov}(Y_t,Y_s) &= (T-t)\wedge(T-s) - (T-t)\wedge T - (T-s)\wedge T + T. \end{aligned} \(If $s \le t$, this simplifies to\) (T-t) - (T-t) - (T-s) + T = s. \(Similarly, if $t \le s$, the result is $t$. Hence\) \operatorname{Cov}(Y_t,Y_s) = \min(s,t). $$
Conclusion.
The covariance structure of $(Y_t)$ matches that of standard Brownian motion.
Key Takeaways.
- Brownian covariance: $\operatorname{Cov}(B_s,B_t)=s\wedge t$.
- Always expand covariance linearly before simplifying.
- Time-reversed increments often collapse cleanly after cancellation.
Part (b) Equality in distribution of processes
Claim.
\(\{Y_t, 0 \le t \le T\} \stackrel{d}{=} \{B_t, 0 \le t \le T\}.\)
Proof.
First,
\(E[Y_t] = E[B_{T-t} - B_T] = 0 - 0 = 0 = E[B_t].\)
From part (a),
\(\operatorname{Cov}(Y_s,Y_t) = s \wedge t = \operatorname{Cov}(B_s,B_t).\)
The process $(Y_t)$ is Gaussian since it is a linear transformation of the Gaussian vector
$(B_{T-t}, B_T)$. A centered Gaussian process is completely determined by its covariance function.
Therefore, $(Y_t)$ and $(B_t)$ have the same finite-dimensional distributions, hence the same
distribution as processes on $[0,T]$.
Conclusion.
$(Y_t)$ is a standard Brownian motion in distribution.
Key Takeaways.
- Linear transformations of Gaussian vectors remain Gaussian.
- For Gaussian processes, mean + covariance completely determine the law.
- “Equality in distribution of processes” means equality of all finite-dimensional distributions.
Part (c)(i) Pathwise identity for the maximum
Claim.
\(Y_T^* = B_T^* - B_T \quad \text{a.s.}\)
Proof.
By definition,
\(Y_T^* = \max_{0 \le s \le T} Y_s
= \max_{0 \le s \le T} (B_{T-s} - B_T).\)
Since $B_T$ is constant in $s$,
\(Y_T^* = \left(\max_{0 \le s \le T} B_{T-s}\right) - B_T.\)
Let $u = T-s$. As $s$ runs from $0$ to $T$, $u$ runs from $T$ down to $0$, so
\(\max_{0 \le s \le T} B_{T-s} = \max_{0 \le u \le T} B_u = B_T^*.\)
Therefore,
\(Y_T^* = B_T^* - B_T \quad \text{a.s.}\)
Conclusion.
The maximum of the time-reversed process has a simple pathwise relationship to the original maximum.
Key Takeaways.
- Maxima commute with time-reversal when paths are continuous.
- Be careful with change-of-variables in maxima: track bounds explicitly.
- This part is almost entirely deterministic once paths are fixed.
Part (c)(ii) Distributional identity via reflection
Claim.
\(|B_T| \stackrel{d}{=} B_T^* - B_T.\)
Proof.
From part (b), $(Y_t){0 \le t \le T} \stackrel{d}{=} (B_t){0 \le t \le T}$.
Since Brownian motion has continuous sample paths, the maximum over $[0,T]$ is determined by the path,
and hence
\(Y_T^* \stackrel{d}{=} B_T^*.\)
From part (c)(i), we have the almost sure identity
\(Y_T^* = B_T^* - B_T,\)
so
\(B_T^* - B_T \stackrel{d}{=} B_T^*.\)
By the reflection principle,
\(B_T^* \stackrel{d}{=} |B_T|.\)
Combining these gives
\(B_T^* - B_T \stackrel{d}{=} |B_T|.\)
Conclusion.
The reflected maximum minus the terminal value has the same law as the absolute value of Brownian motion.
Key Takeaways.
-
Reflection principle: $B_T^* \stackrel{d}{=} B_T $. - Continuity of paths allows extrema to inherit equality in distribution.
- Distinguish carefully between almost sure identities and distributional equalities.
Master Key Takeaways
- Time-reversed Brownian motion (properly centered) is Brownian motion in distribution.
- For Gaussian processes, matching covariance and mean is enough.
- Almost sure identities can be combined with distributional identities to produce new laws.
- Continuity is the silent hypothesis that justifies passing from finite-dimensional distributions to maxima.
Cheat Sheet Entries to Extract
- Time-reversal trick: $B_{T-t}-B_T$ has the same law as $B_t$ on $[0,T]$.
- Gaussian process ID: Mean $=0$ + covariance $= s\wedge t$ ⇒ standard Brownian motion.
-
Reflection principle: $B_t^* \stackrel{d}{=} B_t $. - Maxima + continuity: For continuous processes, equality in distribution passes to suprema.
Notes on My Original Work
- Strong: Correct covariance computation and correct high-level strategy throughout.
- Needed patching: Clarifying the Gaussian-process argument in part (b).
- Main fix: Explicitly invoking continuity (rather than abstract “functionals”) when passing to maxima.
-
Symbol errors in the handwritten version (e.g., $ B_T^* $) were corrected for clarity.
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